For each polynomial function, identify its graph from choices ...

Question

For each polynomial function, identify its graph from choices A–F. ƒ(x)=(x-2)²(x-5)

Six graphs labeled A to F show different polynomial curves with x-intercepts at 2 and 5, illustrating various shapes and turning points.

Accepted Answer

Hello, everybody. I hope you're doing all right. Today, today we're going to be looking at this question that states consider the given polynomial function and plot its graph where our function is F of X is equal to open parentheses, X minus 17, closed parentheses, squared, multiplying and open parentheses. X minus 29 closed parentheses. Now when looking at this question, I see that we have two parts of CS multiplying each other with X terms inside of them. So it's kind of like a factored form already. So when I see that my first instinct is going to be to solve for the zeros of my function and to do that, I just grab whatever was inside of each parentheses and set each one of those equal to zero. So I'll have X minus 17 equals zero and X minus 29 equals zero. Now, X minus 17 equals zero. And we just move the 17 over from the left hand side to the right hand side by adding, which means X is equal to 17 M for X minus 29. We do the same thing. Move it to the right hand side, move the 29 over to the right hand side by adding, so we have X is equal to 29 which means we'll have a point at comma zero and another point at 29 comma zero. And that will function as our zeros. Now for my Y intercept of the graph, all I have to do is set my X equal to zero. So I have F of zero is equal to open parentheses, zero minus 17, closed parentheses, squared, multiplying open parentheses, zero minus 29 closed parentheses. And once we do that and we distribute everything, we'll have a negative 8308 1 or the Y intercept at zero comma negative 381. No, I can look at the different different multiplicities of my zeros. So they each have a different multiplicity first, I'm going to look at X equal 17. So X equals 17 has multiplicity of two. And to look at the multiplicity, you just look at the exponent that each of that X value was raised to. So X equal 17 is part of the uh parentheses that was X minus 17 which was raised to the second power. So we have a multiplicity of two and two is an even number, which means we will be, or the, the graph will be tangent and change directions. Now, for X equals that has multiplicity of one, which is an odd number. Therefore, we will cross the X axis. So those are the multiplicities and what they tell us. Now, finally, I want to see what each end of my graph will look like. And to do that, I just have to look at what type of, what will be the dominant exponent in my function. So if I look at my function, each parentheses has an X inside of it. Now, one of the parentheses is raised to the second power so that X will be raised to the second power. So we'll have an X squared. Now, however, those parties are multiplying so that X squared will be multiplying another XX to give us an X cubed. So end of graph will be a positive X cubed. Therefore, the left hand side will be facing downward and the right hand side of the graph will be facing upwards. And we gather that just from knowing that we have a dominant X cubed exponent that will determine the shape of a graph. So now that I have all that information, I can try to draw a general ship over graph. So I'll draw the Y axis and the X axis and we know we have a point at X equals and another point X equals 29. So we would have 17 comma zero and 29 comma zero as well. And we said we had a why intercept at a very low value of negative 8381. So that isn't winds up zero, comma negative 8381. And now we look at how to connect these dots. So we know each end of my graph. So we know the left hand side is facing downwards. So from the point at X equal 17, from 170.17 comma zero. So the Y intercept, we, we're moving downwards and continuing downwards forever. And from the point at 29 Colin zero, we know we're facing upwards. So from 29 column zero, we go up and now we have both ends of our graphs and now we're missing the area between 17 comma zero and nine and 29 comma zero. So we look at the multiplicities. So at 17 comma zero, we have a multiple 02 which is even and we said we had a tangent here. So we don't cross the X axis. So once we reach 17 comma zero, we touch the X axis change direction and go downwards. So from the Y intercept to the 17 comma zero, we go up, then we touch the X axis change direction, go down and we know at some point the graph is going to have to change direction once again to be able to connect it to the other point at 29 comma zero. And we know at 29 column zero, we have a multiplicity of one which means we cross the X axis. So we cross the X axis at 29 column zero after we flip directions once again. So to recap, we have a point A Y intercept at zero comma negative 8381. So we go from that upwards to a point at 17 comma zero, touch the X axis and change direction back down and then change directions once again upwards to touch the X axis once again at 29 comma zero. And there we are able to cross the X axis and continue infinitely upwards. So that is a general idea of what our graph should look like. So I hope that was helpful. And until next time.

For each polynomial function, identify its graph from choices A–F. ƒ(x)=(x-2)²(x-5)

Key Concepts

Polynomial Functions and Their Graphs

Zeros and Their Multiplicities

End Behavior of Polynomial Functions

Watch next

For each polynomial function, identify its graph from choices A–F. ƒ(x)=(x-2)2(x-5)

Key Concepts

Polynomial Functions and Their Graphs

Zeros and Their Multiplicities

End Behavior of Polynomial Functions

Watch next

For each polynomial function, identify its graph from choices A–F. ƒ(x)=(x-2)²(x-5)